Question

If $2\left[\begin{array}{cc}3& 4\\ 5& x\end{array}\right]+\left[\begin{array}{cc}1& y\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right]$, find $(x-y).$

Answer 1

Consider $2\left[\begin{array}{cc}3& 4\\ 5& x\end{array}\right]+\left[\begin{array}{cc}1& y\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right]$ as shown below:

$\left[\begin{array}{cc}6& 8\\ 10& 2x\end{array}\right]+\left[\begin{array}{cc}1& y\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right]$

$\left[\begin{array}{cc}6+1& 8+y\\ 10+0& 2x+1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right]$

$\left[\begin{array}{cc}7& 8+y\\ 10& 2x+1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right]$

Equating the matrices we get the values of $x$ and $y$ as shown below:

$8+y=0⟹y=-8$

$2x+1=5⟹2x=4⟹x=2$

Hence, $x-y=2-(-8)=2+8=10$.